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设Sn是等差数列an的前n项和,s6/s12=3/10,则s3/s9=¿
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设Sn是等差数列an的前n项和,s6/s12=3/10,则s3/s9=¿
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答案和解析
设首项为a1,公差为d
s6/s12=3/10,
s6/s12=(6a1+15d)/(12a1+66d)
=3(2a1+5d)/6(2a1+11d)
=(2a1+5d)/2(2a1+11d)
则(2a1+5d)/2(2a1+11d)=3/10
解得a1=2d
则s3/s9=(3a1+3d)/(9a1+36d)
=(6d+3d)/(18d+36d)
=9/54
=1/6
s6/s12=3/10,
s6/s12=(6a1+15d)/(12a1+66d)
=3(2a1+5d)/6(2a1+11d)
=(2a1+5d)/2(2a1+11d)
则(2a1+5d)/2(2a1+11d)=3/10
解得a1=2d
则s3/s9=(3a1+3d)/(9a1+36d)
=(6d+3d)/(18d+36d)
=9/54
=1/6
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